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SCOTT HUGHES: Good afternoon.
So we spent our last lecture laying out
some of the basic foundations, making a couple of definitions.
I want to quickly recap the most important concepts
and definitions.
And then, let me be blunt, I kind of
want to get through these definitions, which
I think it's important to do them precisely,
but there's nothing significantly challenging
about them.
We just need to make sure they are defined very precisely.
So now that you've kind of seen the style of these things,
I would like to sort of move through the next batch
of these definitions quickly enough
that we can start to move into more interesting material.
So a quick recap--
and my apologies, my daughter has
some kind of a virus that I am desperately trying to make sure
I do not catch.
And so I will be hydrating during this lecture.
So I just want to recap some of the most important concepts we
went over.
So this whole class is essentially
a study in spacetime.
Later, we're going to connect spacetime to gravity.
And general relativity is going to become
the relativistic theory of gravity.
So we began with a fairly mathematical definition.
Spacetime is a manifold of events
that is endowed with a metric.
Manifold, for our purposes, is essentially
just a set in which we understand
how different members of the set are connected to each other.
Events are really just when and where something happens.
We haven't precisely defined metric yet.
We will soon.
But intuitively, just regard it as some kind
of a mathematical object that gives me a notion of distance
between these events.
I tried and I will continue to try
to be very careful to make a distinction
between geometric objects that live
in the manifold, the events themselves
with things like a displacement vector between two
events, other vectors, which I will
recap the definition of in just a moment.
They have an existence and sort of a reality to them
that is deeper and more fundamental
than the representation of that object.
So when I say that the displacement vector is delta t,
delta x, delta y, delta z, that is according to observer O.
And when I write that down--
I will be very careful as much as possible.
I will occasionally screw this up--
but I will try to write this down
without using an equal sign.
Equal sign implies a degree of reality
that I do not want to impart to that representation.
So an equal dot with a little o here,
that's my personal notation for delta
x is represented by these components
according to observer O. And for shorthand, I was sometimes just
write this by the collection of indices delta x alpha.
A different observer O bar, they will
represent this this factor using the exact same geometric
object.
They all agree that it's this displacement between two
physical events in spacetime.
But they assign different coordinates to it.
They give a different representation to it.
And we find that representation using a Lorenz transformation.
And I'm not going to write out explicitly
the Lorenz transformation matrix lambda.
I gave it in the last lecture.
And I'm assuming you're all experts in special relativity,
and I don't need to go over that.
So I will then using this as sort
of the prototype, the general notion of a vector
in spacetime, which we'll often call
a 4-vector for the obvious reason
that it has 4 components, I'm going
to treat that as any quartet of numbers that has transformation
properties just like the displacement factor
I just went over.
So if there's some quantity A that
has time like an x or y and z component,
as long as a different observer, for reasons having
to do with the underlying physics
or whatever the heck this A is, as
long as that different observer relates their components
to observer O's components via the Lorenz transformation,
you got yourself a 4-vector.
Any random set of four numbers, that ain't a 4-vector.
You need to have some physics associated with it.
And the physics has to tell it that it's
a thing that's related by a Lorenz transformation.
So I want to pick up this discussion
by introducing four particularly important
and special vectors, which to be honest,
we're not going to use too much beyond some
of the first couple of weeks or so of the course,
but they're very useful.
And one should often bear in mind, even later in the course
when they've sort of disappeared,
that they're coming along for the ride secretly.
And these are basis vectors.
So if I go into frame O--
I'm just going to pick some particular reference frame--
I can immediately write down four special vectors.
So remember, this is a Cartesian type of coordinate system.
So I'm going to introduce e0.
I'm going to represent this by just the number 1
in the time slot and 0 everywhere else.
e1, or ex if you prefer, I'm going
to write that down like so.
And you can kind of see where I'm going with this.
The analogy, if you've all seen unit vectors in other classes,
hopefully this is fairly obvious what I'm doing.
I'm just picking out a set of, you know,
little dimensionless simple quantities that point along
the preferred directions that I've
set up in this inertial reference frame.
A compact way of writing this--
so notice, I have four of these vectors.
And these vectors each have four components.
And so what I can say is that the beta component of unit
vector e alpha is representative,
according to observer O, by delta alpha beta, the Kronecker
delta.
If you're not familiar with this one,
I'm usually reluctant to send you to Wikipedia.
But in this case, I'm going to send you
to Wikipedia, Wiki via Kronecker delta.
Yeah, so what that does is it just emphasizes that
at least in-- by the way, I should
put little o's underneath all of these things, because I
have chosen this according to observer o's representation.
So this is just saying that a coin to observer O, these
are four very special vectors.
The utility of these things--
up high for everyone in back to be able to see--
the utility of doing this is that if I now
want to write the vector A as a geometric object,
I can combine the components that observer O uses
with the basis vectors that observer O uses.
And I can sum them all together.
I can put them together.
And then I've got the--
it's not a representation.
That's a damn vector.
OK?
I put it all together using sort of an internally consistent set
of numbers and basis vectors.
And so I am free to say this, where I actually
use an actual equal sign.
You might stop and think, well, but those
aren't the components of the observer O bar would use.
And you're right.
Observer O bar would not use those components.
They would also not use those basis factors.
We're going to talk about how those things change.
But observer O bar does agree that if they were handed
O's components and O's basis vectors,
this would give me a complete representation of what
the vector is.
OK?
So again, I'm really harping on this sort
of distinction between the geometric object
and the representation.
This is the geometric object.
And we can take advantage of this.
The fact that that combination of things
is the geometric object is a tool that I'm going to now--
is a fact that I'm going to exploit in order
to figure out how my basis vectors transform
when I change reference frames.
So let me just repeat what I wrote of there.
But-- oops-- good point for just a slight editorial comment.
When you're talking and writing and there
are millions of little sub scripts and indices,
sometimes the brain and the appendages
get out of sync with one another.
I caught that one.
I don't always.
OK?
So if you see something like that
and you kind of go, um, why did that alpha magically turn
into a beta, it's probably a mistake.
Please call it out.
OK?
All right, so this is how I build this geometric object,
using the components and the basis
vectors as measured by O, as used by observer O.
Let's now write out what they would
be if they were measured by a different observer, an O bar
observer.
I know how to get components, the barred components
from the unbarred components.
I don't yet know how to get the barred basis
vectors from the unbarred basis vectors.
But I know that they exist.
And that once I know what they are, this equation is true.
OK?
These are just two different ways
of writing this geometric object, which
every observer agrees has its existence that
transcends the representation.
So let's rewrite what I've got on the right-hand side
of the rightmost equal sign here.
I'm going to write this as lambda mu bar beta A beta
e mu bar.
And then now I'm going to use a trick, which
is what the-- it's not even so much a trick,
but I'm just going to use a fact that is great when you're
working in index notation.
So often when students first encounter
this kind of notation, your instinct
is to try to write everything out using
matrices and things like row vectors and column vectors.
It's a natural thing to do.
I urge you to get over that.
If I get some bandwidth for that,
I'm going to write up a set of notes
this weekend showing how one can translate at least 2
by 2 objects--
two index objects and one index objects
in a consistent way between matrices and row
vectors and column vectors.
But we're rapidly going to start getting into objects that
are bigger than that, for which trying to represent them
in matrix-like form gets untenable.
In a little while we're going to have a three index object.
And since we don't have three-dimensional chalkboards,
making this sort of matrix representation of that
is challenging.
Soon after that, we'll have four index objects.
And we'll occasionally need to take derivatives
of that four index object, giving us a five index object.
At that point, the ability to sort of treat them
like matrices is hopeless.
So really, you should just be thinking
of this as ordinary multiplication
of the numbers that are represented by these components
as written out here.
And an ordinary multiplication like this,
I can just go ahead and swap the order of multiplication
very easily here.
So what I'm going to do is move the--
hang on a second.
Yeah, yeah, yeah, now I see what I'm doing.
Sorry about that.
I misread my notes.
I'm just going to move the A onto the other side
of my lambda.
And then I'm going to use the fact
that beta on my right-hand side is a dummy index.
So in that final expression I wrote down over there,
beta is a dummy index.
And I'm free to adjust it to put it into a form that's
more convenient for me.
So let's begin.
Let me just write where I've got this equation right now
over there.
So A alpha E alpha equals--
and what I've got then is component A beta lambda mu bar
beta--
keep mu bar.
Sorry about that.
So I'm going to remap my dummy index.
The reason I did that is I can now move this to the other side
and factor out the A alpha.
OK, everyone can see that I hope.
So see what I did was I arranged this
so that I've isolated essentially
only the transformation of the basis vector.
So this equation has to hold no matter
what vector I am working with.
It's got to hold for an arbitrary A alpha.
So the only way that can happen, this
means that my transformation of basis vectors
obeys a law that looks like this.
Now, on first inspection, you're going
to go, ah, that's exactly what we
got for the components of the 4-vector.
Caution-- I'm going to remind you,
it's actually on the board over there.
OK?
The barred component is playing a different role
than the barred basis vector here.
If you want to get the barred basis
vector from the unbarred basis vector,
you need to work with the inverse of this matrix.
That's what this is telling us here.
OK?
All that being said, if you're just working through this
and you've got your components set up
and you're sort of hacking through it,
the algorithm for you to follow is actually simple.
Really, all we're doing is lining up the indices.
We're summing over the ones that are repeated
and requiring that those that are
on both the left-hand side and the right-hand side
appear and equal one another.
Or as an old professor of mine liked
to say about 12 times a lecture, line up the indices.
So that's essentially what we're doing.
In this case, I line up--
if I have my Lorenz transformation
matrix with the barred index up top and the unbarred
down below.
Boom, I line up the indices.
And that tells me what the unbarred basis components
are from the barred ones, and vice versa for the components.
So what basically, as I just said, that tells me
if I actually want to get this guy given this guy,
I need to work with the inverse.
Put this up high so that everyone in the back
can see it.
So this, again, is one of those places where
you might be tempted to sort of write out a matrix
and do a matrix inversion.
But before you do that, remember this is physics.
OK?
The inverse is going to be-- the inverse matrix is going to be
the one that does, at least for certain quantities--
what the Lorenz transformation does
is that it relates objects, if I'm at rest,
if I consider myself at rest, it tells me about things
according to a frame that it's moving with respect to me.
The inverse matrix does the opposite.
It would say to that person being
at rest, what is the matrix that tells them
about things according to me, and they
see me moving with the exact same velocity,
but in the opposite direction.
So to get the inverse matrix, to get the inverse Lorenz
transformation, what we end up doing
is we just reverse the velocity.
So I've just wrote down for you--
and it's worth bearing in mind, every one of these lambdas
that is there, they are really functions.
And so my e alpha that I wrote down over there,
I'm going to use an under tilde to denote a 3 vector.
In one of your textbooks, it's written with a boldface.
But that's hard to do on a blackboard.
So if I want to go the other direction,
well, I just need to have the inverse transformation.
And I have that.
Bear in mind, I mean, those are really the exact same matrices.
OK?
In terms of the function that I'm working with here,
I'm just flipping around the direction in order
to get these things out.
OK?
So, as I said, you might be tempted just
to go ahead and do the matrix inversion.
Let we just do a quick calculation
to show you that that would work.
And the reason I'm doing this is I
just want to quickly step through a particular step,
which is, again, sort of in the spirit of swatting mosquitoes
with sledgehammers, it's the kind of analysis
that you're going to sort of have to do off to the side
now and again.
So given that I know e alpha is--
let's see, let's use beta--
e beta bar be e beta bar.
But I now know that I could write this guy as lambda--
let's use a gamma--
beta bar minus v in the gamma direction.
Now, you look at that, and you go, ooh, look,
I'm actually summing over the betas there.
Let's gather my terms a little bit differently.
So notice, I have unbarred basis vectors over here on the left,
unbarred ones over here on the right,
a bunch of junk in these braces here in the middle.
The only way for that to work is if after summing over beta,
that bunch of junk is, in matrix language, we'd say,
it's the identity matrix.
In component notation, we're going
to call it the Kronecker delta.
Just as little aside, if I started
with the barred on the left side and the unbarred
in the right-hand side and did a similar analysis,
it would take you at this point now that you are all fully
expert in this kind of index manipulation,
it should take you no more than about a minute
to demonstrate to yourself that you can get a Kronecker
delta on the barred indices in a similar way,
just by using them in a slightly different order.
OK?
All right, so far all still basically just formulas.
So I'm going to start now doing a little bit of formalism that
will very quickly segue into physics.
We can all take a deep sigh of relief
and go, ah, OK, something you can imagine measuring.
So that's nice.
So I have introduced 4-vectors.
I haven't introduced the basis vectors and their components.
I haven't really done anything with them yet.
So before we start doing things with them,
let's think about some operations
that we can do with 4-vectors.
So the first one which I'd like to introduce
is a scalar product.
To motivate the scalar product that I'm
going to define in the same way that I defined 4-vectors
as a quartet of numbers whose transformation properties
are based on the transformation properties of the displacement,
I'm going to motivate a general scalar
product between 4-vectors by a similar kind of quantity that
is constructed from the displacement.
So let me recall a result that I hope
is familiar from special relativity.
So working in units where the speed of light is 1,
you are hopefully all familiar with the fact
that this quantity is something that is
in variance to all observers.
I did not use a represented by symbol
there, because no matter whose delta t, delta
x, delta y, delta z I use there, they
will all agree on the delta s squared that comes out of this.
If I want to--
actually, let me write of a few things done
before I say anything more.
So this is an invariant.
This is the same--
let me actually write this in a slightly different way-- it is
the same in all Lorenz frames.
So pick some observer, get their delta t, delta x, et cetera,
assemble delta s squared, pick a different observer,
do the Lorenz transformation, assemble their delta t prime,
delta x prime, et cetera, make that, boom, they all agree.
So we're going to use this to say,
you know what, I'm going to call that the inner product
of the displacement vector with itself.
So I'm going to call this delta x dotted into delta x.
And so what this means is that built into my scalar product--
so if I write this as a particular observer
would compute it--
this is the scalar product that I'm
going to define with respect to the displacement vector.
And this is usually the point where somebody in the class
is thinking, why is there a minus sign
in front of the timelike piece?
I can't answer that.
All I can say is that appears to be--
more than appears to be.
There's a whole frickin' butt load of evidence-- that's
not how nature is assembled.
OK, it's connected to the fact-- so the fact
that all of my spatial directions
are sort of entered with the same sign,
but my timelike direction, it has a different sign.
It's connected to the fact that I can easily move
left and right, front and back.
It's a little bit of effort.
I can move up and down.
But I cannot say, oh, crap, I left my phone at home I'll go
back 15 minutes and pick it up.
You cannot move back and forth in time.
Time actually, which is the timelike component
of this thing, it enters into the geometry
in a fundamentally different way from the spatial things.
And that's reflected in that minus sign.
Anything deeper than that, let's just
say that we're probably not likely to get very
far with that conversation.
Depending on what kind of muscle relaxants
you enjoy using on the weekend, you
might have some fun conversations about it.
But it is not something that you're really
going to get very far with.
You just kind of have to accept that it's
part of the built-in geometry of nature.
OK, so this I am defining as the inner products
of the displacement vector with itself.
I define vectors as having the same transformation properties
as the displacement vector.
We can similarly define an inner product of a 4-vector
with itself.
OK, now, I'll put this on another board.
So A dot A, I'm going to define this--
or rather I will say it is represented according
to observer O as minus A0 squared plus A1 squared plus A2
squared plus A3 squared.
I realize there can be a little bit of ambiguity
in the way I'm writing it here.
You just have to--
if you're ever confused, just ask for clarification
about whether I'm writing something to a power
or whether it's an index label.
Context usually makes it clear.
Handwriting sometimes obscures context though.
The reason why I'm doing this and a real benefit of this
is that whatever this quantity is,
because A has the same transformation
properties of the displacement, this must
be a Lorenz invariant as well.
The underlying mathematics of Lorenz transformation
doesn't care that I wrote a zero instead of delta x here.
It doesn't care that I wrote A1 instead of delta x1, et cetera.
It just knows that it's a thing that
goes into that slot of the Lorenz transformation.
So all observers-- this is how I represent it
according to observer O. But all observers agree on that form.
And as a consequence, this is going to be something
that we exploit a lot.
Even when we move beyond the simple geometry
of special relativity, a generalization of this
will be extremely important for, not an exaggeration to say,
everything.
So a little bit of terminology--
so if A dot A is negative, and depending upon which is bigger,
A0 squared or the sum of the other ones--
it could very well be negative--
we say that A is a time like vector.
This traces back to the fact that if A
were the displacement vector, if the displacement,
the invariant interval, were negative that would tell me
that the two events, which are the beginning and the end
of the interval, I could find a frame at which they're
at the exact same location and are only separated in time.
In the same way, this is basically
saying that I can find a frame when
this vector points parallel to some observer's time axis.
If this is positive, A is spacelike.
Everything I just said about timelike, lather, rinse,
repeat, but replace time with space.
OK?
And if A dot A equals 0, we say A is--
there's two words that are commonly used--
lightlike or null.
Null just traces back obviously to the zero.
Lightlike is because this is a vector that could lie tangent
to the trajectory that a light beam follows in spacetime.
OK?
OK, let me get some clean chalk.
So, so far, I've only talked about this inner product,
this scalar product.
Oh, and, by the way, I'll use inner product and scalar
products somewhat interchangeably.
But this allows me to reiterate a point
I made in Tuesday's lecture.
When I say scalar, scalar refers to a quantity which is--
you know, it doesn't have any components associated with it.
So in that sense, it's familiar from your eager intuition of,
you know, not to vector.
But it has a deeper meaning in this course,
because I also want it to be something
that is invariant between reference frames.
So A dot A is the scalar product.
It gives me a quantity that all observers agree on.
Now, I've only done scalar products
of vectors, A and the displacement vector,
with themselves.
So a more general notion, if I have vectors A and B,
then I will define the inner product between them,
as observed by O, as constructed by observer O, rather, like so.
It's not hard to convince yourself,
given everything we've done so far,
that this quantity must also be invariant.
I'll sketch a really quick proof.
Let's define-- let's say we have two 4-vectors, A
and B. Their sum, by the linearity rules that
apply to these vectors, must also be a vector.
And so if I compute, C dot C, this is an invariant.
With a little bit of labor, that basically
boils down to middle school algebra.
You can show that C dot C is A dot A plus B dot B
plus twice A dot B. This is invariant.
And so this must be invariant.
So this is really useful for us, because we now have a way--
I've introduced these objects, these geometric objects,
these 4-vectors.
We are going to use them in this class
to describe quantities that are of interest
to the physicist who wants to make measurements in spacetime.
We've now learned one of the things
when you're doing stuff in relativity
is you have to be careful who is measuring what.
What are the components that 4-vector as seen
by this observer?
What about their friend who's jogging through the room at 3/4
quarters speed of light?
What about their friend who's driving 2/3 the speed of light
in the other direction?
You have all these really annoying calculations that you
can and sometimes have to do.
This gives us a way to get certain things that
are invariant out of the situation
that everyone is going to agree on.
Invariants are our friends.
So earlier today, earlier in today's lecture,
I talked about how I can write my 4-vectors using the basis
factors.
So another way of writing this--
so what's sort of annoying is every time I've actually
written out the inner product, I have used the represented
by symbol.
I don't want that.
I want to have equal symbols in there, dammit.
So let's take advantage of the fact that A dot B,
I know how to expand A and B using components and basis
vectors.
And again, using the index notation,
I can just pull everything out and rearrange
this a little bit.
Whenever you get down to a point like this,
we now get to do what every mathematician loves to do--
give something a name.
I'm going to define the inner product of basis vector
A with basis vector B to be a two index tensor--
eta alpha beta.
What's lovely about this, this is a totally frame invariant
quantity.
We know that.
And so I've now found a way to write this using the components
as something that gives me a result that is totally
frame invariant.
Now, when you hack through a little bit
the algebra of this, what you'll find
is that the components of this metric--
oh, shoot, I didn't want to actually say it out loud--
the components of this tensor, which pretend
you didn't hear me say that--
the components of this tensor has the following components--
I just said something circular-- has the following values.
This is, as I unfortunately gave away
the plot, this is, in fact, the metric
that I said at the beginning is the quantity that I must
associate with spacetime in order
for there to be a notion of distance between events.
I haven't really said what a tensor is carefully yet.
I'm going to make a more formal definition of this
in just a moment.
But this is your first example of one.
And so the way in which this actually
gives me a notion of distance is through this
that I wrote down right here.
If I have two events in spacetime
that are separated by a displacement delta x,
the delta s squared, which I obtained from this thing,
is fundamentally the notion of distance between those two
events that I use.
And notice, it's a little less normal of a distance
than you're used to when you do sort
of ordinary Euclidean geometry.
This is a distance whose square can be negative.
What we like to say is that when you're
working in special relativity, it's
not necessarily positive-- the distance between two events
is not necessarily-- the distance squared is not
necessarily positive definite.
If it's negative, though, that just means
it's sort of dominated by the time interval between them.
If it's positive, you know it's dominated by the space
interval between them.
If it's zero, well, you actually know--
it's actually a little bit confusing at that point.
They could be, in fact, you know,
very widely separated in both space and time,
but in such a way that a light beam could connect them.
So there's a lot of information encoded in that.
Now, as we move forward--
hang on a second--
as we move forward, we're going to upgrade this.
So right now, our metric is just this simple matrix
of minus 1s, 0s, and 1s.
One of things that we're going to do
is sort of a warm-up exercise to the more complicated things
we're going to do later is we're going
to move away from special relativity
and Cartesian coordinates.
We're going to look at it in polar coordinates.
That's going to be kind of like a warm-up zone.
And so when we do that, we're always
going to reserve eta for the metric of special relativity
when I'm working in Cartesian coordinates.
It's just a great symbol to have for that.
And it's a useful thing to always
have that definition in mind.
I can continue to do special relativity,
but then working in coordinates that are, you know,
spherical-like or polar-like or something like that,
then this is going to become a function.
And what that is going to mean is
that things like my little basis vector
is going to have more complicated behavior.
A little later in the course, we will then
show that when gravity enters into the picture,
essentially the essence of gravity
is going to be encoded in this thing as well in a way
where, again, it's going to be a function.
It's going to vary as a function of space and time.
And the dynamics of gravity will be buried in that.
It's sort of funny that it really does just sort of start
out--
I mean, if you take that thing and you set delta t equals 0,
this is just the bloody Pythagorean theorem.
That is all this is.
Put time back in, and it sort of is the generalization
of Pythagoras to spacetime.
And, in fact, we're going to take advantage of that
and sort of define a geometry that
looks like this as being flat in the same way
that a board is flat, and the Pythagorean theorem works
perfectly on it.
Then, we're going to start think about what
happens when it becomes curved.
And you start thinking about things
like, what is the geometry on the surface of the sphere look
like?
That's just sort of pointing ahead.
So I just throw that at you, so that you
get ready for some of the concepts
that we'll be talking about soon.
So let me write this actually in terms of differentials.
It's sort of useful for what I want to say next.
So a little differential, if I have two events in spacetime
that are very close to one another,
I can write them like so.
And what I've essentially done here
is written dx as dx alpha e alpha.
Before I get into some more sort of a couple
of important, fairly important 4-vectors,
the reason I did this is I want to make an important point
about some notation and terminology that is used.
If it is the case that the displacement vector is related
to the differentials of your coordinates like so,
we say that e alpha is a coordinate basis vector.
What it does is it transforms a differential of your coordinate
into a differential vector in spacetime.
Now, you, may be thinking to yourself, OK, well, what
other kind can there be?
Well, this is where my little spiel there
a second ago about how we're going
to start looking at more complicated things,
it's going to become important.
So when we're working in a Cartesian-like coordinate
system, the fact that this is what we call a coordinate basis
vector isn't very interesting.
Suppose I was working in some kind of a curvilinear
coordinate system, OK?
Spherical coordinates.
Now, let's just focus on 3-space for a second.
So if I write a sort of analogous equation
in curvilinear coordinates--
OK, so here's the 3-space version of that.
Now let's imagine that i equals 1 corresponds to radius,
i equals 2 is theta, i equals 3 is phi.
Then, this would be dr er plus d theta e theta plus d phi e phi.
Does that disturb you at all?
OK, this has dimensions of length.
These have the dimensions of angle.
In order for this to work, er must be dimensionless.
e theta must have the dimensions of length.
e phi must have the dimensions of length.
This is what a coordinate basis looks
like when I am dealing with--
well, we're going to use is a lot in this thing.
I introduce this right now because you are all probably
looking at that, and some small part of you
inside is weeping, because what you want me to write down
is this.
Ah, isn't that better?
OK, this looks like something you're used to.
So I throw this out here right now
just because I want to make sure you're
aware that there are some equations
and some foundational stuff you guys have been doing
over the years, particularly, this shows up a lot when you've
done E&M out of a textbook like Purcell or Griffiths
or Jackson, because there's some derivative operators, which
are assuming that your basis vectors are what we call
orthonormal.
So my e i hat here, it is an orthonormal basis.
And orthonormal basis is defined such
that the dot product of any two members of this thing
gives me back the chronic or delta.
That is not necessarily the case when
I work with a coordinate basis.
Our basis has er dot er equals 1.
Yay, that one's nice.
But when I do e theta dot e theta, I get r squared.
e phi dot e phi will be r squared sine squared theta.
And what I'm going to do when I start
generalizing these things, I'm going to change my--
this thing which I defined up here--
do I still have it on the board?
Yeah, yeah, right here.
So when I set eta alpha beta is e alpha dot e beta and I made
it this thing, I'm going to generalize this and say
at the dot product of any two basis
vectors it gives me a more general notion
of a metric tensor.
And the values in the metric tensor
maybe functions like this.
Right now, throw that out there, you know,
this might be sort of just like a peak of the horrors
that lie ahead.
OK, we're not going to worry about this too much just yet.
But I want you to be prepared for this.
In particular, it's really useful
to have this notion of a coordinate basis
versus an orthonormal basis in your head.
We're going to start defining some derivative operations
soon.
In fact, probably won't get to them today,
but they will be present when we start doing Lecture 3.
And there's a couple of results that come up
where everyone's sort of like, wait, I
knew that the divergence had a factor of r on that derivative
there.
Where'd it go?
It's because we're not working in an orthonormal basis.
All right, I'm a little sick of math.
So let's do a little physics.
So, so far, the actual only physical 4-vector
that I've introduced is the displacement vector.
From the displacement vector, it's really easy
to make the probably the first and simplest
important 4-vector, which is known as the 4-velocity.
This tells me the rate of displacement of an observer
as this person moves through spacetime per unit--
and we're going to be careful about this in this class--
d tau is the time interval as measured
along the trajectory of the observer with 4-velocity u.
In other words-- that's a very long winded way of saying it--
it's an interval of proper time.
In English, the word proper time sounds very like, woo,
I don't want to use improper time.
I better use that.
But this actually I think it comes from French.
It just refers to the fact that it's one's own time.
Apparently in German people say eigenzeit.
So, you know, there's a couple of different words for it.
But proper time is what we use.
In special relativity, if we see someone
going by with constant velocity, a particular observer who sees,
you know--
we're here in the room.
Someone comes through.
Their 4-velocity is u.
We would see their 4-velocity to have the components
gamma gamma v, where gamma, I'll remind you,
is the special relativistic Lorenz factor.
And I'll remind you again we've set speed of light to 1.
A very useful thing, which we're actually
going to take advantage of quite a bit,
is that in the rest frame of u--
pardon me for a second--
in the rest frame of u, or I should
say of the observer whose 4-velocity is u,
they just have 1 in atomic direction, C,
if you want to put your factors back into their.
And that's basically just saying that the person is standing
still, but moving through time, because you are always
moving through time.
All right, from the 4-velocity for an observer who has--
or for an object, I should say, who has some mass,
we can easily define the 4-momentum,
where this m is known as the rest mass of this object.
It's worth a bit of description here.
You will often see, particularly in some older textbooks that
discuss special relativity, people
like to talk about the relativistic mass.
And that comes from the fact that if I write out
what this thing looks like according
to some particular observer, you have this gamma m entering
into both of the components.
And so older textbooks often called
gamma m the relativistic mass.
That's not really the way people have--
over the course of the past couple decades,
they've moved away from that.
And it's just more useful to focus on the rest mass
as the only really meaningful mass,
because, as we'll see in a moment,
it's a Lorenz invariant.
We'll see how that is in literally about 3 minutes.
And so what we're instead going to say
is that as seen by some particular observer,
this has a timelike component that
is the energy that that observer would measure
and a set of spacelike components
that are the momentum that that observer will measure.
So where we get a bit of important physics out of all
this stuff is by coupling these two 4-vectors to the scalar
products that we made up.
So the first one, if you do u dot u,
according to any observer, that's
just going to be minus gamma squared
plus gamma squared v squared.
And with about 20 seconds worth of analysis,
you can find that this is always equal to minus 1.
Actually, there's an even trickier way to do this.
Suppose I evaluate this in the rest frame of the observer
whose 4-velocity is u?
Well, in the rest frame, v is 0, and gamma is 1,
and I get minus 1.
And this is an invariant.
So whatever I get in that particular frame
must be obtained in all frames.
That's a trick we're going to use
over and over and over again.
Sometimes you can identify--
you know, you get some kind of God awful expression
that just makes you want to vomit.
But then you go, wait a minute, what
would this look like in frame blah, blah, blah?
And you sort of think about some particular frame.
And in that frame, it may simplify.
And if it does and it's a frame invariant quantity, mazel tov,
you have just basically won the lottery.
You've got this all taken care of.
Go on with your life.
So the 4-velocity has a scalar product of itself
that is always minus 1.
OK?
How about the 4-momentum?
Well, the 4-momentum is just 4-velocity times mass.
So that's just minus m squared.
But we also know it's related to these two other quantities,
which are important in physics.
This is related to the energy and to the momentum.
So this is also equal to minus e squared plus--
so this is the ordinary 3--
the magnitude of the 3-vector part of this thing as
measured by the observer who breaks up
the 4-momentum in this way.
So what this means is I can manipulate this guy
around a little bit here.
Anybody who works in particle physics
is presumably familiar with this equation.
Sometimes it appears with the p squared
moved onto the other side.
If it looks a little bit unfamiliar to you,
let me put some factors of C back in this.
So remember, we have set C equal to 1.
When you put it back in, that's what this is.
So it drops out of this in a very, very simple way.
One of the uses of this--
and many of you have done exercises presumably
in some previous study that does this.
And there'll be one exercise on the p set
that was just posted where you exploit this.
So a key bit of physics, the reason
why we care about 4-momentum is it's in one mathematical object
allows us to combine conservation energy
and conservation of momentum.
So conservation of 4-momentum puts both conservation
of energy and conservation of momentum
into one mathematical object.
So if I have n particles that interact,
then the total 4-momentum is conserved in the interaction.
OK?
So, yeah, we talk a little bit more about this
and then just sort of quickly move on.
So combining this with the fact that we
are free to change our reference frames often
it gives us a trick that allows us to really simplify
a lot of analysis.
So if I have n particles that are
sort of swarming around and doing
some horrible bit of business that I need to study
and I need to have a good understanding of,
we can often vastly simplify our algebra
by choosing a special and very convenient frame of reference
in which to do our analysis--
choose the center of momentum frame.
So this is the time frame in which that that p tote--
so C-O-M, center of momentum--
has zero spatial momentum.
In that frame, you have just as much momentum going to the left
as going to the right, just as much going forward
as going backwards, as much going up as going down.
And so this turns out to be-- so the classic example of where
this is really useful is when you are studying particle
collisions and you're looking at things like the production
of new particles.
So imagine you've got particle A with some 4-momentum PA coming
in like this.
Particle B's got some 4-momentum coming in like this.
These guys collide.
And they do so--
I work in the center of momentum frame.
I might want to just calculate the energy at which they just
happen to produce some new pair of particles at rest.
I would like to find the threshold
for this particular creation process.
OK?
So you're going to play with one problem
on the p-set that's kind of like that.
Let's see, what do I have time to do?
I think I will do--
yeah, I think I can do two more things.
So all the dot products that I have been talking about so far
have been a dot product of a 4-vector with itself.
I did u dotted into u.
I did p dotted into p.
I invented a frame in which p has a particularly simple form.
And then when you actually do some of analysis,
you would probably take that p tote and dot it into itself.
I haven't done anything that looks
at the crossing between these two things,
dotting one into the other.
So let me to go through a very useful result that
follows by combining p with u.
There's a very specific notion of p
and a very specific notion of u.
Let's let p be the 4-momentum of a particle.
I call it a.
Let's let u be the 4-velocity not of a, but the 4-velocity
of observer O. So particle a might
be a muon that was created the upper atmosphere
and is crashing through our room right now.
Observer O might be your hyperactive friend
who is jogging through the room at half the speed of light.
The question I want to ask is, what does O measure
as the energy of particle a.
So the naive way to do this, which I will emphasize
is not wrong, what you might do is sort of go,
OK, well, we're sitting here.
This room is our laboratory.
I've measure this thing in my lab.
So I know p as I measure it.
I can see O jogging by.
So I know O's 4-velocity as I measure it.
So what I should do is figure out the Lorenz transformation
that takes me into the rest frame of O.
Once I have that Lorenz transformation,
I'll apply that Lorenz transformation
to the 4-vector p.
Boom, that will give me that energy.
That will work.
That will absolutely work.
But there's an easier way to do it.
So one thing you should note is that everybody represents that
4-velocity as an energy--
excuse me, they represent the 4-momentum
as an energy and a 3-momentum.
In particular, though, they represent it
as the energy that they would measure
and the 3-momentum that they would measure.
So that means p as seen by O is e according to O
and p according to O. That are acquainted o is what we want.
And I just told you a moment ago,
you know, if you have p in your own reference frame,
and you have u in your own reference frame,
you can do this whole math with Lorenz transformations
and get it out.
But you also know that in O's own reference frame,
O represents their 4-velocity as 1 in the timelike direction, 0
in the spatial direction.
So what this means is if I go into O's reference frame,
if I go into their inertial reference frame,
notice that if I take the dot product of p and u,
I get e times 1 and p times 0.
So that is just negative.
It's exactly what I want-- modulo minus sign.
And so you go, OK, well, I'll flip my minus sign around.
And you think, OK, great, but I did
that using those quantities as written down
in O's reference frame.
And then you go, holy crap, that's
the invariant scalar product.
I'm done.
Mic drop.
Leave the room.
What this means is you start with p as you measure it,
u as you measure it.
Take the scalar product between the two of them.
Boom, the answer you want pops out.
No nonsense of Lorenz transformations.
None of that garbage needs to happen.
You just take that inner product and you've got it.
So that sort of says in words, no matter
what representation you choose to write down p and u in,
take the dot product between the two of them,
throw in a minus sign, you've got the energy of the particle
with p as measured by the observer with u.
It's sort of late in the hour, or the hour and a half
I should say.
So let me just sort of emphasize,
there are occasional moments in this class where
if you're dozing off a little bit, I suggest you pop up
and tattoo this into a neuron somewhere.
This is one of those moments.
OK?
This is a result that we're going to use over and over
and over again, because this holds--
this isn't just in special relativity.
When we start talking about the behavior of things
near black holes, there's going to a place
where I basically at that point to say,
well, I'm going to use the fact that the observer measures
an energy that is given by-- and I'm going to write down that.
The dot product that's involved is a little bit more
complicated, because my metric is hairier.
But it's the exact same physical concept.
OK?
Let me just do one more.
And then I'll go talk, without getting into the math,
about what I will start with on Tuesday.
Last 4-velocity, which is probably useful for us
to quickly talk about is--
so we've talked a lot about 4-velocity.
That is just one piece-- when we're
talking about sort of the kinematics of bodies moving
in spacetime, you need more information than just velocity.
Sometimes things are moving around.
There's additional forces acting on them.
And so we also care about the 4-acceleration.
And so this is what I get when I take
the derivative with respect to proper time of the 4-velocity.
So will there be some homework exercises that use this.
The main thing which I want to emphasize to sort of conclude
our calculations for today is that when
I talk about a 4-velocity of acceleration,
this has an extremely important property.
It is always the case that a dotted into u equals 0.
If you're used to sort of 3-dimensional intuition,
that may seem weird.
OK?
Anytime you see a car accelerate from a stop, that's
a case in which its acceleration is clearly not
orthogonal to its velocity.
But the issue here is, this is not a spatial dot product.
This is a space time dot product.
And some of your intuition has to go out the window
because of that.
It's very simple to prove this.
Remember, u dot u equals minus 1.
So d of u dot u, d tau, which is just 2, u
dot a is the derivative of minus 1, which is 0.
OK?
So this is something that we will exploit.
If you want to describe the relativistic kinetics
of an accelerating body, this is a great thing
that we can use to exploit.
You often need a little bit more information.
We have to give you as a bit of additional information
some knowledge about what the orientation of the acceleration
is and things like that.
So whenever you are given--
I'll get to you in just a sec-- whenever you're
given any kind of differential quantity like this,
it's not enough to know-- it's like the acceleration of loss,
you have to also have boundary conditions.
And that sort of tells you what the initial direction is.
Question?
AUDIENCE: Is that time still the proper time?
SCOTT HUGHES: That time is the proper time, yes.
AUDIENCE: So it's not accelerating?
SCOTT HUGHES: That's right.
So you can still define a proper time for an accelerating
observer.
It will not relate--
hold that thought.
You're going to play this a little bit more
in a future problem set.
I mean, the key thing is that the way--
if you have an d observer, an interval of proper time
as compared to an interval of time for, you know, someone
in a rest frame that sees this person accelerate away,
the conversion between the intervals of time,
the two measure, it evolves.
So, you know, let's say you're in this room with me.
In fact, it turns out that if you accelerate
at 1G for a year, you get to very
close to the speed of light.
So let's say that you were in a rocket ship
right now that launched with an acceleration of G. Initially,
you and I synchronize our watches.
And so an interval of a second to me
is the same as a second to you.
Half a year later, you're moving at something
like half the speed of light.
And I will see a noticeable time delay.
An interval of a second as you measure it
looks long compared to me.
Six months later, you're actually
quite close to the speed of light,
and it gets dilated even more.
So last thing, which I'm going to say,
and I'm not going to get into too much detail with this yet,
is we're going to begin next time by making
a little bit more formal some of the notions that go around.
So we've increased some physics and some vectors.
I've given you guys one tensor so far, the metric tensor.
And so I'm going to give you-- in fact,
I will write down a very precise definition of this right now,
and we'll pick it up from there on Tuesday.
So the basic idea-- so you guys have see--
the one tensor you've seen so far is the metric tensor.
And what the metric is is it's sort of a mathematical object
that I put in a pair of 4-vectors.
And it spits out a quantity that is a Lorenz invariant
scalar that characterizes what we call the inner product
of those two 4-vectors.
More generally, I'm going to define
a tensor of type 0 N as a function or mapping of N
4-vectors into Lorenz invariant scalars, which is linear
in each of its N arguments.
So I will pick it up here on Tuesday.
And let me just say in words, the metric is a 0 2 tensor.
I put in two 4-vectors.
It spits out a Lorenz invariant scalar.
We're going to before too long come up
with a couple of things that involve three real vectors--
excuse me three 4-vectors, too many numbers here,
a trio of 4-vectors, which it then maps to a Lorenz invariant
scalar.
Some of them will take in four 4-vectors
and produce a Lorenz invariant scalar.
Notice I wrote this in sort of funny way.
The 0 N sort of begs for there to be sort of an N 0.
OK?
To do that I have to introduce an object that
is sort of dual to a vector.
We're going to talk about that.
Those are objects called one-forms,
which actually happen to be a species of vector.
We're actually going to then learn that the vector is itself
a tensor.
And so we will make a very general classification
of these things.
And we'll see that vectors are just a subset of these tensors.
And at last, we'll sort of have all the mathematics in place.
We can sort of lose some of these distinctions
and just life goes on, and we can start actually
doing some physics with these.
All right, I will pick it up there on Tuesday.